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12:00 AM
@TedShifrin Who, me?
I have a graph. You assign a probability distribution to the vertices - each vertex gets a probability between 0 and 1 inclusive, and they add up to one.
I will choose two vertices independently at random according to your probability distribution.
You win if those vertices are distinct and joined by an edge. I win if the vertices are either equal, or are distinct and not joined by an edge.
(a) What is the formula for your probability of winning, given the probability distribution? (b) How do you maximize your chance of winning?
12:32 AM
@AkivaWeinberger I would like to use the K-distribution here
$f_X(x; b, v)= \frac{2b}{\Gamma(v)} \left( \sqrt{bx} \right)^{v-1} K_{v-1} (2 \sqrt{bx} )$
1 hour later…
1:53 AM
@geocalc33 No idea what that is but I don't think it's right
4 hours later…
5:57 AM
I want to show that given a manifold with the maximal atlas $\{(U_a, \phi_a)\}$, and a point $p\in M$ with a neighborhood $U\ni p$, there exists a $U_b$ such that $p\in U_b\subset U$.
I think the following works: Since the atlas is maximal, there is some a such that $p\in U_a\implies p\in U_a\cap U.$ Define $g=\phi_a|_{U_a\cap U}$. For any $b,$ if $U_b\cap U_a\cap U$ is empty, then the charts $(U_a\cap U, g)$ and $(U_b,\phi_b)$ are compatible . Suppose the intersection to be non empty. Then $g\circ \phi_b^{-1}$ is $C^\infty$ at any $x$ in the intersection (because $g$ is restriction of $\ph
any objections to this attempt of the proof?
@Koro its fine - i would give full marks to someone who said it follows because inclusion maps from open subsets of manifolds are smooth
or in other words, differentiability at a point is a local property at that point
6:43 AM
thanks :-)
2 hours later…
8:34 AM
Instead of taking algebra 2 course, I decided to take number theory course. Probably I regret my decision a lot, but whatever.
wait, I decided not to take any number theory class because of the shock of hensel's lemma in elementary number theory class. Interesting.
8:56 AM
@MatsGranvik isn't it more about the limits
9:13 AM
How to construct an example of a continuous function $f:[0,\infty)\to R$ such that $\lim_{t\to \infty} \int_{[0, t]} f d\lambda$ exists but the integral $\int_R f d\lambda, \lambda$ is Lebesgue measure does not.?
Why $||f_n||_{L^2}\leq 1$ implies ${f_n(x)\over n}\to 0$ as $n\to\infty$ for a.e. $x$?
Clearly $(0, 0,0) \in D$
@onepotatotwopotato $|f_n|(x) \le 1$ for a.e. x.
I have to find a path from $(0, 0,0) $ to $(a, b, c) \in D$
9:24 AM
@Koro hmm... why?
@Koro think about the alternating series test and why it works.
@Koro this should give you ample intuition to construct such a function
then what you want is for $\lim_{t \rightarrow \infty} \int_{0}^t f d\lambda$ to converge because of the alternating series test, but you want the lebesgue integral to not exist, because both the positive and negative series diverge.
so for probably the easiest concrete case of this happening, use the harmonic series, so that $1 - \frac{1}{2} + \frac{1}{3} .... $ converges, the positive series is $1 + \frac{1}{3} + \frac{1}{5} + ... $ and the negative series $-\frac{1}{2} - \frac{1}{4} - ...$, both of which by comparison to the harmonic series, diverge
The other day I was trying to construct a sequence of functions $f_n$ such that $f_k(x)\to 0$ as $k\to \infty$, and $\lim_{k\to \infty}\int f_k=\infty$.
I was thinking of triangles with increasing height and it doesn't work...
you can actually turn this into a concrete example of a function $f$ satisfying your requirements (i leave the details to you)
One of my friends looked at it and said: take $f_k(x)= 1/k$...
it worked. I don't know why it didn't occur to me then 😅
well, the harmonic series is basically the go to for these kinds of things, because its the boundary point at which $\frac{1}{n^s}$ series converge/diverge
9:30 AM
@porridgemathematics hmm thinking...
btw @onepotatotwopotato just because the L^2 norm is less than or equal to one, does not give you that inequality there..
as in |f_n(x)|<=1 a.e. is not true just because |f_n|_{L^2|<=1
It's stated in Stein's RA so I wondered why
you mean your original question
because the function constantly equal to 10 on [0,1/100] and zero elsewhere has L^2 norm equal to 1
but is certainly not less than 1 a.e. in absolute value.
Oh you mean $|f_n(x)|\leq 1$. I didn't think it's true.
yeah just wanted to point out that is definitely not true
your original question im still thinking about (its probably true by virtue of sum 1/n^2 converging or something..)
what are the other hypothesis, what measure space are you on etc
or is it completely general measure space
9:36 AM
Anyway my original statement is in Stein's book saying 'we saw that ...'
I didn't see that before.
Nothing. It's just stated like that: 'We saw that if $||f_n||_{L^2}\leq 1$, then ${f_n(x)\over n}\to 0$ as $n\to\infty$ for a.e. $x$.'
ohhh, maybe try looking at f_n^2/n^2
sum _{n=1}^infinity |f_n|^2/n^2 is in L^2
uh i mean L^1
by comparison with sum_{n=1}^infinite 1/n^2
and the triangle inequality
therefore \sum_{n=1}^infinity |f_n|^2/n^2 is finite for a.e. x
i.e. is convergent for a.e. x
hence lim_n |f_n|/n = 0 for a.e. x
i think that works @onepotatotwopotato
(something like that should work at least)
It seems correct to me too. Thanks
@porridgemathematics ohh
@onepotatotwopotato nvm
@Koro yeah its a nice thing to remember for these types of problems
replacing integral with sum can be helpful
(it turned out both of your problems were examples where thinking about sums helped)
10:16 AM
@Koro by the way, for your most recent question, there is a simple way to directly see why that union is open. What happens when $||x||$ is very small? You should be able to see that you automatically get that when this is the case, those points lie in the second set besides $\{0 \}$, so that your second set actually contains a small punctured disk at $0$
so you can certainly proceed directly, using your argument
@porridgemathematics I don't understand how to construct the example yet.
use a riemann sum
this uses complex integral.
@porridgemathematics for what function?
on $[0,1]$ set your function to $1$, on $[1,2]$ to $-\frac{1}{2}$, and so on
10:19 AM
its really just the sum
you barely have to work any harder
you can convince yourself that the limit equals 1 - 1/2 + blah
but the lebesgue integral doesnt exist
because the positive and negative series are also just sums
which diverge
i think this is a better example than those answers
but yeah the accepted one is a common example too
its the dirichlet integral
which is useful in fourier analysis
of course there will be jumps in my case
but you can smoothen my function out at the jumps
and get it as smooth as you like
I understand that the limit exists.
but I don't yet understand why Lebesgue integral does not exist.
I got it.
$f= f^+- f^-$
$\int_0^\infty f^+=\infty=\int_0^\infty f^-=\infty$
so the Lebesgue integral is not defined.
this is cooked up precisely for this question
@porridgemathematics sum and integrals are so closely related.
but you can do it with any conditionally convergent series
try and prove it
that is, you can get infinitely many counterexamples, by choosing the values of the function to be equal to the sequential values of conditionally convergent series
whose positive series and negative series need to both diverge
(otherwise they wouldnt be conditionally convergent in the sense im talking about)
yes, for conditionally cgt. series, either the series of positive or negative terms will diverge.
10:26 AM
both will for me
otherwise technically you can rearrange the terms
and the thing will still blow up
so im allowing +infinity
and -infinity as limits
as long as they are well defined limits
meaning no matter how you order the summation
you converge to the same thing in $[-\infty,+\infty]$
in that case, a series is only conditionally convergent when both its positive and negative series converge
@porridgemathematics yes
thanks :)
This example is not continuous though but I think that's not a problem as we can smoothen it.
its really not an issue yeah
you can smoothen it by drawing a picture and making sure your smoothenings dont contribute more than finitely to the integral (i.e. the lebesgue integral of the smoothened portions is finite)
so for continuity, its enough to just draw a bunch of straight lines in between the positive and negative ends
with gradient getting steeper and steeper from left to right of the graph
so that your area contributions there are not important
(an easy way is to make their area contributions sum like a geometric series)
10:55 AM
@Koro i tried to answer your question about why the 'direct' method wasnt working (it also works)
i mean your definition method
11:11 AM
@porridgemathematics I suspected that: the union should absorb 0 somewhere to give an open set. Thanks.
np, and yeah thats what has to happen to get openness
Q: In $(X,S,\mu), \int f^r d\mu<\infty\implies \int f^q d\mu<\infty$, where $\mu(X)<\infty$.

KoroSuppose that $(X,S,\mu)$ is a measure space with $\mu(X)<\infty$. Suppose that $f:X\to [0,\infty)$ is $S-$ measurable. Suppose that $q>r$ are given positive numbers. Then, $\int f^rd\mu<\infty\implies \int f^q d\mu<\infty.$ I tried to prove it the following way: For brevity, let $\{f>k\}$ stand f...

I think this is correct. But I posted it to get it vetted.
11:23 AM
If I've a finite field and I want to find the number of primitive elements of the multiplicative group, I have to use the totient function right?
12:17 PM
@SineoftheTime primitive elements?
@SineoftheTime thats right, because you are computing the number of generators of a cyclic group
$f(x)= \sin x/x $ is not Lebesgue integrable on (0,\infty).

Robjohn style answer: $f=f^+-f^-$.

$\begin{align} \int_{[0,\infty)} f^+=&\sum_{k=0}^\infty \int_{2k\pi}^{(2k+1)\pi} \sin x/x\tag 1\\ \ge & \sum_{k=0}^\infty \frac 1{(2k+1)\pi}\int_0^\pi \sin x dx\tag 2 \\=&2\sum_{k=0}^\infty \frac 1{(2k+1)\pi}\gt \infty\tag 3\end{align}$
its interesting that the harmonic series appears again
Q: Textbooks to read after Armstrong Topology

Colin SavageI just finished reading through Armstrong - Topology and I'm looking for a logical next step to keep going in learning more topology. I've taken courses in Real + Complex Analysis, Algebra, Geometry and others and am a third-year undergrad for reference. I'm considering Hatcher - Algebraic Topo...

1:05 PM
@Jakobian Yes the Euler Gamma constant is a limit, the von Mangoldt function is a limit, the natural logarithm is a limit, the logarthmic derivative of the Riemann zeta function is a limit, the Riemann zeta zeros are a limit, the Riemann zeta function is a limit.
b) is definitely true ( by using Vieta's formulas)
Now I want to show that $D$ is path connected.
Is $D$ a linear subspace? Is $D$ is a cone? Is $D$ a convex set?
$(0, 0,0) \in D$
Let $(a, b, c) \in D$
That's why my algebra professor didn't like me.
1:44 PM
@porridgemathematics ty
@Koro yes
@SouravGhosh you dont need anything special to see $D$ is connected
just use that $D$ is the image of a continuous function from $\mathbb{R}^3$
Any general remarks or advice on graduate life you wanna tell? e.g, exercise regularly, sleep enough, etc, or something in academics. @TedShifrin @leslietownes
@SouravGhosh the coefficients are continuous functions of the roots
being connected is a more basic property than being a cone or being a linear subspace or being a convex set. so asking those questions is asking something more specific to get information to assist you with something a lot more basic about $D$
2:06 PM
@porridgemathematics Thanks. Can you write the explicit map?
its just +/- times elementary symmetric functions
i.e. from vietes formulas
ill write it out
(x_1,x_2,x_3) -> (-x_1-x_2-x_3,x_1x_2 + x_1x_3 + x_2x_3 , -x_1x_2x_3)
$(\alpha, \beta, \gamma) \to (-a, b, -c) $
@porridgemathematics Thanks :)
@onepotatotwopotato unsolicited, but spend less time on MSE and general 'procrastaworking' (i would tell myself when I started, lol)
basically ignore free dopamine from answering random peoples questions and answer your own (usually harder dopamine)
2 hours later…
4:18 PM
How would one classify/categorize the symmetries in the GCD(n,k) matrix?
Greatest Common Divisor=GCD
$$ \Delta \varphi = - \frac{xn}{\sum_{i=1}^n t_i} \varphi_x $$
with $\varphi(x,t_1,t_2,\cdot\cdot\cdot,t_n)$
4:38 PM
Hi everyone! Can anyone suggsest me some Mathematical Mahazine/Periodicals, where undergrads are allowed to publish any article explaining a particular topic or get problems published in it,
or solutions to problems in the magazine/journal maybe something like "Crux Mathematicorum." I examined about the "American Mathematical Monthly " but it seems one has to subscribe to that in the website of MAA, to read it. This means, one cannot contribute if one doesn't subscribe, for then one wouldn't be aware what's written there i.e unaware of the contents of it. So, any other popular mathematics based journals/periodicals/magazine if suggested, would be really helpful!
1 hour later…
5:45 PM
@ThomasFinley Romanian Mathematical Magazine is one. Has some of the most insane looking integrals I've ever seen.
6:23 PM
American Math Monthly is probably higher level, anyhow. But the AMS/MAA do have a few journals aimed more at undergraduates, like Mathematics Magazine and College Mathematics Journal.
1 hour later…
7:43 PM
How can I show this: Let $\varphi_0$ be a homomorphism from an integral domain $R$ to algebraically closed field $M$ and $F$ be a field containing $R$. Then $\varphi_0$ extends either to an embedding $\varphi$ of $F$ into $M$ or to a place $\varphi$ of $F$ into $M\cup \{\infty\}$?
7:53 PM
@JaakkoSeppälä I don’t understand your English. Do you have examples.?
I try to write it on the other way: Consider an integral domain $R$ and an algebraically closed field $M$. Let $\varphi_0$ be a homomorphism from $R$ to $M$, and let $F$ be a field that contains $R$. In this context, we can explore two possibilities for extending $\varphi_0$: either as an embedding $\varphi$ of $F$ into $M$, or as a place $\varphi$ of $F$ into $M\cup {\infty}$."
Is it clearer?
Well, I don't have examples, I'm just reading a book.
Place $\varphi$ ?
Into suggests injection, which is wrong, too.
A place of a field F is a map φ of F into a set M ∪ {∞}, where M is a
field, with these properties:
(a) φ(a + b) = φ(a) + φ(b).
(b) φ(ab) = φ(a)φ(b).
(c) There exist a, b ∈ F with φ(a) = ∞ and φ(b) not equal to 0, ∞.
I suggest you always try to understand new statements with examples.
Strange word. So it’s a homomorphism but we have to define addition/mult of infinity.
So try to give an example where you get an embedding and an example where you get a “place” — maybe those examples will give you understanding.
Okay. I try to do that. Thanks!
8:09 PM
@leslie Is today duck pond day?
we went on saturday and saw ten baby ducks.
Wow ... no joke about Make Way for Ducklings.
yeah, it was cute. they even came over to us, and got pretty close, once munchkin realized that they might not flee into the water if she just sat still.
She will learn from ducks, but not from humans.
yes, ducks are a big influence
8:22 PM
Next, it'll be cats.
1 hour later…
9:43 PM
your need an ai duck
@冥王Hades I just looked back at that problem. Did you get $x=48$?
Oh, I see it in the image.
10:04 PM
Howdy @copper and @robjohn.
@TedShifrin good afternoon
10:16 PM
$\phi:R\to S$ surjective ring homomorphism, $\mathfrak b$ ideal of $R$. how come $\phi^{-1}(\phi(\mathfrak b)) = \mathfrak b + \ker \phi$?
i know we should have $\mathfrak b \subseteq \mathfrak b^{ec}$, but where is that equality coming from?
Gads, the day has gotten away from me. It's just over an hour until I take my dog to the park.
Almost the pooch’s dinner time!
As soon as we get back from the park
Then it will be our dinner time, too
I have no idea what your notation is, but just prove it. @shin
shin: anything in b + ker(phi) gets mapped to something in phi(b), right? that's one inclusion. conversely, if z is in phi^(-1)(phi(b)), there's w in b with phi(z) = phi(w), and then z = w + (z - w) exhibits z as a sum of an element (w) of b and an element (z - w) of ker(phi).
10:23 PM
tough love vs spoon feeding
shin: think about where, if at all, phi being surjective might be used in this argument, or outside of this argument, why it would matter generally
ultimately the point is to prove that surjective homomorphisms take prime ideals to primes, maximals to maximals, etc.; there's just this step i'm stuck at. currently reading what you wrote
OK, so there are additional areas where some or all of the hypotheses on phi might come up
Surjectivity is crucial just to get out the door.
@robjohn Are you suggesting my spoon has jagged edges?
A more universal tool: cutting and scooping
10:29 PM
shin: probably worth thinking generally about which pieces of your arguments use/need phi being surjective and b being an ideal, as opposed to just phi being a homomorphism or b being some more general subset of R
dig in and scoop out the heart of the matter
at leslie: on the first direction, we do get that b + ker(phi) gets mapped to something in phi(b), but i'm not seeing how the inverse map restores b. consider Z -> Z_8
with, e.g., the ideal <7>
The whole ring?
$\mathfrak b$ is the ideal of $R$ we're extending and then contracting
maybe the poster did not mean the entire ideal is restored, only its contraction
math.stackexchange.com/a/292857/755010 if you want to give it a go, but i might just go with a simpler proof, not sure where he's getting that equality from and i've been thinking a while
@TedShifrin Happy Memorial Day (albeit war is never a cause for celebration...)
10:42 PM
10:56 PM
oh i understood
i was making a tremendous derp
i woke up at 5 am today and it's late, for some reason i was thinking of inverse maps as mapping an element to an element instead of mapping an element to an ideal
That is wrong.
Did you not learn about preimage of a set in your intro to higher math coursr?
i know about preimages, they're not functions and map a single element to multiple others in case of surjective functions
for some reason brain was not computing
That’s garbled, but this is that in the setting of a homomorphism.
yeah in the case of Z -> Z_8 the preimages of the canonical homomorphism are ideals
11:11 PM
when f is a function from X to Y, some books will actually introduce distinct notation for the "preimage under f" function from the power set of Y to the power set of X, instead of writing f^{-1} for it. in my experience, this turns out to be more annoying than not doing it, and doesn't fix the issue of this being a significant stumbling block for a lot of students.
i had to teach out of a book that did this once. the trauma has erased the book's title from my memory.
That sentence is no good. The preimage of a single element is only an ideal in one case.
at ted: what about $\phi:\mathbb Z \to \mathbb Z_8: z \mapsto z + 8\mathbb Z$, and we take $\phi^{-1}(5+8\mathbb Z)$? don't we get the set containing the solutions to $x \mod 8 = 5$?
at shin: yes. and that's not an ideal (for example, because it doesn't contain 0, which is a problem you might encounter with preimages of many other single elements)
that's why you add $\ker \phi$
cool cool, thanks a lot
11:30 PM
shintuku, I asked you sometime ago but you missed my question. Are you in school or self-studying? I am questioning since Ted once mentioned that you have gaps in your math knowledge.
self-studying, currently econ major

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